Bzoj3530 [Sdoi2014] Count

Time Limit:10 Sec Memory limit:512 MB
submit:744 solved:394 Description

We call a positive integer n is a lucky number when and only if its decimal representation does not contain any element in the number string set S as its substring. For example, when s= (22,333,0233), 233 is the lucky number, 2333, 20233, 3223 is not the lucky number.
Given N and S, calculate the number of lucky numbers not greater than N.

Input

The first line of the input contains the integer n.
The next line is an integer m, which represents the number of elements in S.
The next m line, a number string per line, represents an element in S.

Output

Outputs an integer that represents the value of the answer modulus 109+7.

Sample Input20
3
2
3
14
Sample Output -HINT

The L in the following table represents the length of N, and L represents the sum of all the string lengths in S.

1 < =l < =1200, 1 < =m < =100, 1 < =l < =1500

Source

Round 1 Day 1

DP on AC automaton

Just a little bit more trouble than the DP on the trie tree.

First build the trie tree, set up a good AC automaton, and then run the digital DP. The case where the length of the number is less than n is processed separately, the maximum bit limit is not considered at this time. After processing a number length equal to n length, when the highest bit is not filled to separate the decision.

See the code specifically:

1 /*by Silvern*/2#include <iostream>3#include <algorithm>4#include <cstring>5#include <cstdio>6#include <cmath>7#include <queue>8 using namespacestd;9 Const intmod=1e9+7;Ten intt[1510][ One],cnt=1; One BOOLend[1510];intfail[1510]; A voidInit () { for(intI=0; i<=9; i++) t[0][i]=1;} - voidInsertChars[]) { -     intrt=1, I,j; the     intlen=strlen (s); -      for(i=0; i<len;i++){ -         if(!t[rt][s[i]-'0']) t[rt][s[i]-'0']=++CNT; -rt=t[rt][s[i]-'0']; +     } -end[rt]=1; +     return; A } at intq[15100],hd,tl; - voidBuild () { -Hd=0, tl=1; -q[++hd]=1; -fail[1]=0; -      while(hd<=TL) { in         intnow=q[hd++]; -end[now]|=End[fail[now]]; to          for(intI=0; i<=9; i++){ +             intv=T[now][i]; -             if(!v) { thet[now][i]=T[fail[now]][i]; *             } $             Else{Panax Notoginseng                 intk=Fail[now]; -                  while(!t[k][i]) k=Fail[k]; thefail[v]=T[k][i]; +q[++tl]=v; A             } the         } +     } -     return; $ } $ intn,m; - Chars[1510],c[1510]; - inta[1510]; the intf[1510][1510][2]; - intans=0;Wuyi intMain () { thescanf"%s", s+1); -     inti,j; Wu     intLen=strlen (s+1); -      for(i=1; i<=len;i++) a[i]=s[i]-'0'; About init (); $scanf"%d",&m); -      for(i=1; i<=m;i++){ -scanf"%s", c); - Insert (c); A     } + Build (); the      for(i=1; i<=9; i++)//First -         if(!end[t[1][i]]) $f[1][t[1][i]][0]++; the      for(i=1; i<len-1; i++)  the          for(j=1; j<=cnt;j++) the              for(intk=0; k<=9; k++){ the                 if(!End[t[j][k]]) { -f[i+1][t[j][k]][0]+=f[i][j][0]; inf[i+1][t[j][k]][0]%=MoD; the                 } the             } About      for(i=1; i<len;i++) the       for(j=1; j<=cnt;j++) ans= (ans+f[i][j][0])%MoD; theMemset (F,0,sizeoff); the      for(i=1; i<=a[1];i++) +         if(!end[t[1][i]]) { -             if(i==a[1]) f[1][t[1][i]][1]++; the             Elsef[1][t[1][i]][0]++;Bayi         } the      for(i=1; i<len;i++) the        for(j=1; j<=cnt;j++) -          for(intk=0; k<=9; k++) -             if(!End[t[j][k]]) { thef[i+1][t[j][k]][0]+=f[i][j][0]; thef[i+1][t[j][k]][0]%=MoD; the                 if(k<a[i+1]){ thef[i+1][t[j][k]][0]+=f[i][j][1]; -f[i+1][t[j][k]][0]%=MoD; the                 } the                 Else if(k==a[i+1]){ thef[i+1][t[j][k]][1]+=f[i][j][1];94f[i+1][t[j][k]][1]%=MoD; the                 } the             } the      for(i=1; i<=cnt;i++){98Ans= (ans+f[len][i][0])%MoD; AboutAns= (ans+f[len][i][1])%MoD; -     }101printf"%d\n", ans);102     return 0;103}

Bzoj3530 [Sdoi2014] Count